Let
⎧ -x+b, if x < -3
f(x) = ⎨ 2, if x = -3
⎩ (−2)/(x−b) + 1, if x > -3 (and x≠b)
What I tired is to set -x + b = $(−2)\over(x−b)$ + 1 where x is approaching -3. I get b = $7±\sqrt{33}\over{-2}$. Which is incorrect? please help!
a) For what value(s) of b is f continuous at -3?
b) For what value(s) of b does f have a removable discontinuity at -3?
Why were you trying to set $-x + b = \frac {-2}{x-b} + 1$?? That is completely pointless.
To have $f(x)$ continuous at $-3$, you need to have $$\lim_{x\to -3}f(x) = f(-3)$$ Because $f(x)$ is defined differently on each side of $-3$, it is useful to break the full limit into one-sided limits. So the condition becomes:
$$\lim_{x\to -3-}f(x) = f(-3)\\\lim_{x\to -3+}f(x) = f(-3)$$
Now $f(-3) = 2$, while $\lim_{x\to -3-}f(x)$ only involves the behavior of $f(x)$ for values of $x < -3$ and $\lim_{x\to -3+}f(x)$ only involves the behavior of $f(x)$ for values of $x > -3$.
$f(x)$ has a removable discontinuity at $-3$ if $\lim_{x\to -3}f(x)$ converges, but $\lim_{x\to -3}f(x) \ne f(-3)$. $\lim_{x\to -3}f(x)$ converges if both one-sided limits converge, and are equal. So a removable discontinuity requires
$$\lim_{x\to -3-}f(x) = \lim_{x\to -3+}f(x)$$
Also note that $\lim_{x\to -3-}f(x)$ and $\lim_{x\to -3+}f(x)$ are numbers. When you compute each of them, the value will not involve "$x$" at all (they will involve $b$, since the function $f$ is defined in terms of $b$, but $x$ is only a dummy variable - the value of the limit does not depend on it).